Question

For the chemical reaction, $2O_3\rightleftharpoons 3O_2$
The reactions proceed as follows
$O_3\rightleftharpoons O_2+O$(fast)
$O+O_2\to 2O_2$ (slow)
The rate law expression will be:

• A. $r=k^{\prime }[O_3{]}^2$
• B. $r=k^{\prime }\left[O_3{]}^2\right[O_2{]}^{-1}$
• C. $r=k^{\prime }\left[O_3\right]\left[O_2\right]$
• D. unpredictable

Answer 1

Answer: (B)

$r=k^{\prime }\left[O_3{]}^2\right[O_2{]}^{-1}$

For slow step, rate $=K\left[O\right]\left[O_3\right]$.....(i)
Also for equilibrium $K_c=\frac{\left[O\right]\left[O_2\right]}{\left[O_3\right]}$........(ii)

$\therefore \left[O\right]=K_c\frac{\left[O_3\right]}{\left[O2\right]}$ .......(iii)
By (iii) and (i), the intermediate [O] is eliminated as rate, $\frac{K.Kc\left[O3\right]\left[O3\right]}{\left[O2\right]}=K\text{'}\frac{[O_3{]}^2}{\left[O2\right]}$
Hence b is the correct answer.